A heavy sled is being pulled by two people as shown in the figure. The coefficie
ID: 1431192 • Letter: A
Question
A heavy sled is being pulled by two people as shown in the figure. The coefficient of static friction between the sled and the ground is s = 0.587, and the kinetic friction coefficient is k = 0.467. The combined mass of the sled and its load is m = 366 kg. The ropes are separated by an angle = 23°, and they make an angle = 30.4° with the horizontal. Assuming both ropes pull equally hard, what is the minimum rope tension required to get the sled moving? If this rope tension is maintained after the sled starts moving, what is the sled's acceleration?
Explanation / Answer
The normal force
Fn = mg - 2Fsin = 366kg * 9.8m/s² - 2 * F * sin30.4º
Fn = 3586.8N - 1.012*F
static friction Ff = µ*Fn =0.587 * (3586.8N - 1.012*F ) = 2105.45N - 0.594F
The horizontal component in the direction of motion of the two ropes must overcome the friction:
2 * F * cos * cos(/2) = 2105.45N - 0.594F
1.69F =2105.45N - 0.594F
2.162F =1674.6N
F =921.825 N to get the sled moving
BONUS: what force is required to now KEEP it moving?
kinetic friction Ff = µ*Fn =
0.467 * (3586.8N - 1.012*F ) = 1674.6N - 0.472F
The horizontal component in the direction of motion of the two ropes must overcome the friction:
2 * F * cos * cos(/2) = 1579N - 0.47F
1.69F = 1674.6N - 0.472F
2.162F =1674.6N
F = 774.56 N end BONUS
(I misread the question and did this by mistake!)
The friction WAS Ff = 2105.45N - 0.594*921.825N = 1557.86 N,
so that IS the applied horizontal force.
The friction NOW IS Ff = 0.467 * (3586.8N - 1.012*774.56N ) = 1309 N
so the net force
Fnet = 1557.86N - 1309N = 248.86 N
acceleration a = Fnet / mass = 248.86 N / 366kg = 0.679 m/s²
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